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E n = 1 2 m e v 2 k Zq e 2 r n . size 12{E rSub { size 8{n} } = { {1} over {2} } m rSub { size 8{e} } v rSup { size 8{2} } - k { { ital "Zq" rSub { size 8{e} } rSup { size 8{2} } } over {r rSub { size 8{n} } } } } {}

Now we substitute r n size 12{r rSub { size 8{n} } } {} and v size 12{v} {} from earlier equations into the above expression for energy. Algebraic manipulation yields

E n = Z 2 n 2 E 0 ( n = 1, 2, 3, ... ) size 12{E rSub { size 8{n} } = - { {Z rSup { size 8{2} } } over {n rSup { size 8{2} } } } E rSub { size 8{0} } \( n=1," 2, 3, " "." "." "." \) } {}

for the orbital energies of hydrogen-like atoms    . Here, E 0 size 12{E rSub { size 8{0} } } {} is the ground-state energy n = 1 size 12{ left (n=1 right )} {} for hydrogen Z = 1 size 12{ left (Z=1 right )} {} and is given by

E 0 = 2 π 2 q e 4 m e k 2 h 2 = 13.6 eV.

Thus, for hydrogen,

E n = 13.6 eV n 2 ( n = 1, 2, 3, ...).

[link] shows an energy-level diagram for hydrogen that also illustrates how the various spectral series for hydrogen are related to transitions between energy levels.

An energy level diagram is shown. At the left, there is a vertical arrow showing the energy levels increasing from bottom to top. At the bottom, there is a horizontal line showing the energy levels of Lyman series, n is one. The energy is marked as negative thirteen point six electron volt. Then, in the upper half of the figure, another horizontal line showing Balmer series is shown when the value of n is two. The energy level is labeled as negative three point four zero electron volt. Above it there is another horizontal line showing Paschen series. The energy level is marked as negative one point five one electron volt. Above this line, some more lines are shown in a small area to show energy levels of other values of n.
Energy-level diagram for hydrogen showing the Lyman, Balmer, and Paschen series of transitions. The orbital energies are calculated using the above equation, first derived by Bohr.

Electron total energies are negative, since the electron is bound to the nucleus, analogous to being in a hole without enough kinetic energy to escape. As n size 12{n} {} approaches infinity, the total energy becomes zero. This corresponds to a free electron with no kinetic energy, since r n size 12{r rSub { size 8{n} } } {} gets very large for large n size 12{n} {} , and the electric potential energy thus becomes zero. Thus, 13.6 eV is needed to ionize hydrogen (to go from –13.6 eV to 0, or unbound), an experimentally verified number. Given more energy, the electron becomes unbound with some kinetic energy. For example, giving 15.0 eV to an electron in the ground state of hydrogen strips it from the atom and leaves it with 1.4 eV of kinetic energy.

Finally, let us consider the energy of a photon emitted in a downward transition, given by the equation to be

Δ E = hf = E i E f . size 12{ΔE= ital "hf"=E rSub { size 8{i} } - E rSub { size 8{f} } } {}

Substituting E n = ( 13.6 eV / n 2 ) size 12{E rSub { size 8{n} } = - "13" "." 6``"eV"/n rSup { size 8{2} } } {} , we see that

hf = 13.6 eV 1 n f 2 1 n i 2 . size 12{ ital "hf"= left ("13" "." 6" eV" right ) left ( { {1} over {n rSub { size 8{f} } rSup { size 8{2} } } } - { {1} over {n rSub { size 8{i} } rSup { size 8{2} } } } right )} {}

Dividing both sides of this equation by hc size 12{ ital "hc"} {} gives an expression for 1 / λ size 12{1/λ} {} :

hf hc = f c = 1 λ = 13.6 eV hc 1 n f 2 1 n i 2 . size 12{ { { ital "hf"} over { ital "hc"} } = { {f} over {c} } = { {1} over {λ} } = { { left ("13" "." 6" eV" right )} over { ital "hc"} } left ( { {1} over {n rSub { size 8{f} } rSup { size 8{2} } } } - { {1} over {n rSub { size 8{i} } rSup { size 8{2} } } } right )} {}

It can be shown that

13.6 eV hc = 13.6 eV 1.602 × 10 −19 J/eV 6.626 × 10 −34 J·s 2.998 × 10 8 m/s = 1.097 × 10 7 m –1 = R

is the Rydberg constant    . Thus, we have used Bohr’s assumptions to derive the formula first proposed by Balmer years earlier as a recipe to fit experimental data.

1 λ = R 1 n f 2 1 n i 2 size 12{ { {1} over {λ} } =R left ( { {1} over {n rSub { size 8{f} } rSup { size 8{2} } } } - { {1} over {n rSub { size 8{i} } rSup { size 8{2} } } } right )} {}

We see that Bohr’s theory of the hydrogen atom answers the question as to why this previously known formula describes the hydrogen spectrum. It is because the energy levels are proportional to 1 / n 2 size 12{1/n rSup { size 8{2} } } {} , where n size 12{n} {} is a non-negative integer. A downward transition releases energy, and so n i size 12{n rSub { size 8{i} } } {} must be greater than n f size 12{n rSub { size 8{f} } } {} . The various series are those where the transitions end on a certain level. For the Lyman series, n f = 1 size 12{n rSub { size 8{f} } =1} {} — that is, all the transitions end in the ground state (see also [link] ). For the Balmer series, n f = 2 size 12{n rSub { size 8{f} } =2} {} , or all the transitions end in the first excited state; and so on. What was once a recipe is now based in physics, and something new is emerging—angular momentum is quantized.

Triumphs and limits of the bohr theory

Bohr did what no one had been able to do before. Not only did he explain the spectrum of hydrogen, he correctly calculated the size of the atom from basic physics. Some of his ideas are broadly applicable. Electron orbital energies are quantized in all atoms and molecules. Angular momentum is quantized. The electrons do not spiral into the nucleus, as expected classically (accelerated charges radiate, so that the electron orbits classically would decay quickly, and the electrons would sit on the nucleus—matter would collapse). These are major triumphs.

Questions & Answers

Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
Jyoti Reply
I only see partial conversation and what's the question here!
Crow Reply
what about nanotechnology for water purification
RAW Reply
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
yes that's correct
I think
what is the stm
Brian Reply
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
scanning tunneling microscope
how nano science is used for hydrophobicity
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
what is differents between GO and RGO?
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
analytical skills graphene is prepared to kill any type viruses .
what is Nano technology ?
Bob Reply
write examples of Nano molecule?
The nanotechnology is as new science, to scale nanometric
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Is there any normative that regulates the use of silver nanoparticles?
Damian Reply
what king of growth are you checking .?
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
Stoney Reply
why we need to study biomolecules, molecular biology in nanotechnology?
Adin Reply
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
what school?
biomolecules are e building blocks of every organics and inorganic materials.
anyone know any internet site where one can find nanotechnology papers?
Damian Reply
sciencedirect big data base
Introduction about quantum dots in nanotechnology
Praveena Reply
what does nano mean?
Anassong Reply
nano basically means 10^(-9). nanometer is a unit to measure length.
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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Source:  OpenStax, College physics -- hlca 1104. OpenStax CNX. May 18, 2013 Download for free at http://legacy.cnx.org/content/col11525/1.1
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